Question

Tính giá trị biểu thức sau: 

P=\(\dfrac{1}{1+2}\)+\(\dfrac{1}{1+2+3}\)+\(\dfrac{1}{1+2+3+4}\)+...+\(\dfrac{1}{1+2+3+4+...+2021}\)

Answer 1

Ta có: \(P=\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+\cdots+2021}\)

\(=\frac{1}{2\cdot\frac32}+\frac{1}{3\cdot\frac42}+\cdots+\frac{1}{2021\cdot\frac{2022}{2}}\)

\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\cdots+\frac{2}{2021\cdot2022}\)

\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2021\cdot2022}\right)\)

\(=2\left(\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\right)=2\left(\frac12-\frac{1}{2022}\right)\)

\(=1-\frac{1}{1011}=\frac{1010}{1011}\)